To Show: that f is one-one and onto

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x₁, x₂ ∈ A & f(x₁), f(x₂) ∈ B, f(x₁) = f(x₂) ↔ x₁= x₂ or x₁ ≠ x₂ ↔ f(x₁) ≠ f(x₂)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto.

Let x₁, x₂ Q and f(x) = \(\frac{x-1}{x-2}\).So f(x₁) = f(x₂) → \(\frac{x_1-1}{x_1-2}\) = \(\frac{(x_2-1)}{x_2-2}\), on solving we get x₁=x₂

So f(x₁) = f(x₂) x₁= x₂, f(x) is one-one

Given co-domain of f(x) is R – {1}

Let y = f(x) = \(\frac{x-1}{x-2}\), So x = \(\frac{2y-1}{y-1}\) [Range of f(x) = Domain of y]

So Domain of y = Range of f(x) = R – {1}

Hence, Range of f(x) = co-domain of f(x) = R – {1}.

So, f(x) is onto function

As it is a bijective function. So it is invertible

Invers of f(x) is f^-1(y) = \(\frac{2y-1}{y-1}\)